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Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 Section 6.6 Solving Quadratic Equations by Factoring Copyright © 2013, 2009, 2006 Pearson Education,

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Presentation on theme: "Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 Section 6.6 Solving Quadratic Equations by Factoring Copyright © 2013, 2009, 2006 Pearson Education,"— Presentation transcript:

1 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 Section 6.6 Solving Quadratic Equations by Factoring Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

2 2

3 3 Solving Polynomial Equations We have spent much time on learning how to factor polynomials. Now we will look at one important use of factoring. In this section, we will use factoring to solve equations of degree 2. Up to this point, we have only looked at solving equations of degree one.

4 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 Solving Quadratic Equations Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form where a, b, and c are real numbers, with. A quadratic equation in x is also called a second-degree polynomial equation in x. The Zero-Product Rule If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB = 0, then A = 0 or B = 0.

5 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Solve the equation (x + 5) (2x – 1) = 0. By the zero-product principle, the only way this product can be zero is if at least one of the factors is zero. Thus, x + 5 = 0 or2x – 1 = 0 x = – 5 2x = 1 x = ½ The two solutions are – 5 and ½. Solving Quadratic EquationsEXAMPLE

6 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 Solve the equation (x + 5) (2x – 1) = 0. By the zero-product principle, the only way this product can be zero is if at least one of the factors is zero. Thus, x + 5 = 0 or2x – 1 = 0 x = – 5 2x = 1 x = ½ The two solutions are – 5 and ½. Solving Quadratic EquationsEXAMPLE

7 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 Objective #1: Example

8 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 8 Objective #1: Example

9 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 9

10 10 Solving Polynomial Equations Solving a Quadratic Equation by Factoring 1) If necessary, rewrite the equation in the standard form, moving all terms to one side, thereby obtaining zero on the other side. 2) Factor. 3) Apply the zero-product principle, setting each factor equal to zero. 4) Solve the equations formed in step 3. 5) Check the solutions in the original equation.

11 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 11 Solve x 2 – 7x = 8 Step 1: Rewrite the equation in standard form ax 2 + bx + c = 0. x 2 – 7x = 8 x 2 – 7x – 8 = 0 Step 2: Factor: x 2 – 7x – 8 = 0 (x – 8) (x +1) = 0 Solving Quadratic Equations by FactoringEXAMPLE

12 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 12 Solve x 2 – 7x = 8 Step 1: Rewrite the equation in standard form ax 2 + bx + c = 0. x 2 – 7x = 8 x 2 – 7x – 8 = 0 Step 2: Factor: x 2 – 7x – 8 = 0 (x – 8) (x +1) = 0 Solving Quadratic Equations by FactoringEXAMPLE

13 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13 Solve x 2 – 7x = 8 by factoring Steps 3 and 4: Set each factor equal to zero and solve. x – 8 = 0 or x + 1 = 0 x = 8 or x = –1 Step 5: Check the solutions in the original equation. 8 2 – 7(8) = 8(–1) 2 – 7(–1) = 8 64 – 56 = = 8 8 = 8 8 = 8 Solving Quadratic Equations by FactoringCONTINUED

14 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Solve x 2 – 7x = 8 by factoring Steps 3 and 4: Set each factor equal to zero and solve. x – 8 = 0 or x + 1 = 0 x = 8 or x = –1 Step 5: Check the solutions in the original equation. 8 2 – 7(8) = 8(–1) 2 – 7(–1) = 8 64 – 56 = = 8 8 = 8 8 = 8 Solving Quadratic Equations by FactoringCONTINUED

15 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 45 from both sides and write the equation in standard form. Subtract 45 from both sides Simplify 2) Factor. Factor

16 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. ? ? ? ?

17 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Solving Polynomial EquationsCONTINUED The solutions are 9 and – 5. The solution set is {9, – 5}. ?? The graph of is shown here. true

18 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 4x from both sides and write the equation in standard form. Note: Do not divide both sides by x, or we would lose a potential solution. Subtract 4x from both sides Simplify 2) Factor. Factor

19 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 0: Check 4: ?? true

20 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Solving Polynomial EquationsCONTINUED The solutions are 0 and 4. The solution set is {0,4}. The graph of is shown here.

21 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: Be careful! Although the left side of the original equation is factored, we cannot use the zero-product principle since the right side of the equation is NOT ZERO!! 1) Move all terms to one side and obtain zero on the other side. Subtract 14 from both sides and write the equation in standard form. Simplify Subtract 14 from both sides

22 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 Solving Polynomial EquationsCONTINUED 2) Factor. Before we can factor the equation, we must simplify it first. FOIL Simplify Now we can factor the polynomial equation.

23 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 23 Solving Polynomial EquationsCONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. or 5) Check the solutions in the original equation. ?? ??

24 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Solving Polynomial EquationsCONTINUED true The solutions are 3 and – 6. The solution set is {3, – 6}. The graph of is shown here.

25 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 25 Objective #2: Example

26 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 26 Objective #2: Example

27 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 27 Objective #2: Example

28 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Objective #2: Example

29 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Objective #2: ExampleCONTINUED

30 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30 Objective #2: ExampleCONTINUED

31 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Objective #2: Example

32 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 32 Objective #2: Example

33 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 33 Objective #2: ExampleCONTINUED

34 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 34 Objective #2: ExampleCONTINUED

35 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 35 Objective #2: Example

36 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 36 Objective #2: Example

37 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 37 Objective #2: ExampleCONTINUED

38 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 38

39 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 39 Polynomial Equations in ApplicationEXAMPLE A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The function describes the height of the gymnast’s feet above the ground, s (t), in feet, t seconds after dismounting. The graph of the function is shown below.

40 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 40 Polynomial Equations in ApplicationSOLUTION When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph. We note that the graph of the equation passes through the line y = 8 twice. Once when x = 0 and once when x = 0.5. This can be verified by determining when y = s (t) = 8. That is, CONTINUED Original equation Replace s (t) with 8 Subtract 8 from both sides Factor

41 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 41 Polynomial Equations in Application Now we set each factor equal to zero. CONTINUED We have just verified the information we deduced from the graph. That is, the gymnast will indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5 seconds. These solutions are identified by the dots on the graph on the next slide. or

42 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 42 Polynomial Equations in ApplicationCONTINUED

43 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 43 Objective #3: Example

44 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 44 Objective #3: Example

45 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 45 Objective #3: ExampleCONTINUED

46 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 46 Objective #3: ExampleCONTINUED


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